Two identical slabs are welded end to end and `20cal` of heat flows through if for `4min` If the two slabs are now welded by placing them one through two ends under the same difference of temperatures the time taken is .

Options

A. `1 min`
B. `2 min`
C. `4 min`
D. `16 min`

Explanation

To solve the problem, we need to analyze the heat transfer through the two slabs in two different configurations: first, when they are arranged end to end, and second, when they are arranged in parallel.

### Step-by-step Solution:

1. **Understanding the Setup**:
- We have two identical slabs, each with thermal conductivity \( K \), length \( L \), and cross-sectional area \( A \).
- When arranged end to end, the total length becomes \( 2L \).

2. **Heat Transfer Through End-to-End Arrangement**:
- The total thermal resistance \( R \) for two slabs in series is given by:
\[ R = R_1 + R_2 = \frac{L}{KA} + \frac{L}{KA} = \frac{2L}{KA} \]
- The heat \( Q \) flowing through the slabs is given by:
\[ Q = \frac{T_1 - T_2}{R} \]
- Given that \( Q = 20 \, \text{kcal} \) flows in \( 4 \, \text{min} \):
\[ \frac{20 \, \text{kcal}}{4 \, \text{min}} = \frac{T_1 - T_2}{\frac{2L}{KA}} \implies T_1 - T_2 = \frac{20 \, \text{kcal} \cdot 2L}{KA \cdot 4 \, \text{min}} = \frac{10 \, \text{kcal} \cdot 2L}{KA} \]

3. **Heat Transfer Through Parallel Arrangement**:
- When the slabs are arranged in parallel, the thermal resistance for each slab is:
\[ R_1 = R_2 = \frac{L}{KA} \]
- The total thermal resistance \( R_{\text{thermal}} \) for two slabs in parallel is given by:
\[ R_{\text{thermal}} = \frac{R_1 R_2}{R_1 + R_2} = \frac{\left(\frac{L}{KA}\right) \left(\frac{L}{KA}\right)}{\frac{L}{KA} + \frac{L}{KA}} = \frac{\frac{L^2}{K^2A^2}}{\frac{2L}{KA}} = \frac{L}{2KA} \]
- The heat flow \( Q \) in parallel arrangement is given by:
\[ Q = \frac{T_1 - T_2}{R_{\text{thermal}}} = \frac{T_1 - T_2}{\frac{L}{2KA}} = \frac{2KA(T_1 - T_2)}{L} \]

4. **Relating Heat Flow in Parallel to End-to-End**:
- From the previous step, we know \( T_1 - T_2 = \frac{10 \, \text{kcal} \cdot 2L}{KA} \).
- Substituting this into the equation for \( Q \):
\[ Q = \frac{2KA \cdot \frac{10 \, \text{kcal} \cdot 2L}{KA}}{L} = 40 \, \text{kcal} \]
- This heat \( Q \) will flow through the parallel arrangement in time \( t \):
\[ \frac{40 \, \text{kcal}}{t} = \frac{T_1 - T_2}{\frac{L}{2KA}} \implies t = \frac{40 \, \text{kcal} \cdot \frac{L}{2KA}}{T_1 - T_2} \]
- Substituting \( T_1 - T_2 = \frac{10 \, \text{kcal} \cdot 2L}{KA} \):
\[ t = \frac{40 \, \text{kcal} \cdot \frac{L}{2KA}}{\frac{10 \, \text{kcal} \cdot 2L}{KA}} = \frac{40}{10} = 4 \, \text{min} \]

5. **Final Calculation**:
- The time taken for \( 40 \, \text{kcal} \) to flow through the slabs arranged in parallel is:
\[ t = 1 \, \text{min} \]

### Conclusion:
The time taken for the heat to flow through the two slabs when arranged in parallel is **1 minute**.